2log5 x+log5 100-2=0

2 log (5) x+log (5) 100-2=0; x>0

2log (5) x+2 log (5) 10-2=0

log (5) х+log (5) 10-1=0

log (5) x+log (5) 2+log (5) 5-1=0

log (5) x+log (5) 2=0

log (5) x=-log (5) 2

ĺog (5) x=log (5) 2^ (-1)

x=1/2

2log5 (x) + log5 (100) - 2 = 0

log5 (x^2) + log5 (100) = 2

log5 (x^2*100) = 2

x^2*100 = 25

x^2 = 0,25

x = ±0,5

ОДЗ х>0, поэтому х = 0,5